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May $Z$ be a random variable distributed as $N(0,1)$ Find the following limit: \begin{equation} \lim_{n\to\infty}\mathbb{E}\left[\frac{1}{\sqrt{n}-Z}\right] \end{equation}

How does one go about proving it?

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up vote 1 down vote accepted

The expected value does not exist: the function $ \dfrac{f(z)}{\sqrt{n}-z}$ (where $f$ is the probability density function) is not absolutely integrable because of the singularity at $z=\sqrt{n}$. However, the Cauchy principal value of this improper integral does exist. That is, if $I_r(t) = 1$ for $|t|\ge r$ and $0$ for $|t|<r$, ${\mathbb E}\left[ \dfrac{I_r(\sqrt{n}-Z)}{\sqrt{n}-Z} \right]$ exists, and for the limit of this as $r \to 0$ I get $\sqrt{\dfrac{\pi}{2}} e^{-n/2} \text{erfi}(\sqrt{n/2})$, which goes to $0$ as $n \to \infty$.

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Thanks, V.P. of the integral is what one wants in this case. –  Alex Lomachenko Dec 13 '12 at 13:41
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